%I #10 Sep 11 2017 05:49:12
%S 1,1,-1,1,-3,2,1,-9,14,-6,1,-45,110,-90,24,1,-585,1670,-1710,744,-120,
%T 1,-21105,61670,-66150,32424,-7560,720,1,-1858185,5439350,-5864670,
%U 2925384,-728280,91440,-5040,1,-367958745,1077215510,-1161894510,580489224,-145567800,18961200,-1285200,40320,1
%N Triangle T, read by rows, equal to the matrix inverse of the triangle defined by [T^-1](n,k) = A075263(n,k)/n!, for n>=k>=0.
%C Row sums are zero after the initial row. Absolute row sums equal A106339.
%F Also, T(n, k) = k!*A106340(n, k), where A106340 is the matrix inverse of the triangle formed from (n-k)!*A008278(n, k), n>=k>=0 and A008278 is the triangle of Stirling numbers of 2nd kind.
%e Triangle begins:
%e 1;
%e 1,-1;
%e 1,-3,2;
%e 1,-9,14,-6;
%e 1,-45,110,-90,24;
%e 1,-585,1670,-1710,744,-120;
%e 1,-21105,61670,-66150,32424,-7560,720;
%e 1,-1858185,5439350,-5864670,2925384,-728280,91440,-5040; ...
%e The matrix inverse T^-1 begins:
%e 1;
%e 1,1;
%e 1,3/2,1/2;
%e 1,2,7/6,1/6;
%e 1,5/2,25/12,5/8,1/24;
%e 1,3,13/4,3/2,31/120,1/120;
%e 1,7/2,14/3,35/12,301/360,7/80,1/720; ...
%e where [T^-1](n,k) = A075263(n,k)/n!.
%e Each row n of the matrix inverse equals the initial
%e (n+1) fractional coefficients of (x/log(1+x))^n,
%e which are listed below for n=1,2,3,...,9:
%e 1; 1/2,-1/12,1/24,-19/720,3/160,-863/60480,275/24192,...
%e 1,1; 1/12,0,-1/240,1/240,-221/60480,19/6048,...
%e 1,3/2,1/2; 0,1/240,-1/480,1/945,-11/20160,47/172800,...
%e 1,2,7/6,1/6; -1/720,0,1/3024,-1/3024,199/725760,...
%e 1,5/2,25/12,5/8,1/24; 0,-1/6048,1/12096,-19/725760,...
%e 1,3,13/4,3/2,31/120,1/120; 1/30240,0,-1/57600,1/57600,...
%e 1,7/2,14/3,35/12,301/360,7/80,1/720; 0,1/172800,...
%e 1,4,19/3,5,81/40,23/60,127/5040,1/5040; -1/1209600,0,...
%e 1,9/2,33/4,63/8,331/80,37/32,605/4032,17/2688,1/40320; 0,...
%t rows = 10; Tinv = Table[(1/n!)*PadRight[CoefficientList[x^(n+1)*Sum[k^n * (1-x)^k, {k, 0, Infinity}], x], rows], {n, 0, rows-1}]; T = Inverse[Tinv ]; Table[T[[n, k]], {n, 1, rows}, {k, 1, n}] // Flatten (* _Jean-François Alcover_, Sep 11 2017 *)
%o (PARI) T(n,k)=(M=matrix(n+1,n+1,m,j,if(m>=j, polcoeff((-x/log(1-x+x^2*O(x^n)))^m,j-1)))^-1)[n+1,k+1]
%o (PARI) T(n,k)=(-1)^n*k!*(matrix(n+1,n+1,r,c,if(r>=c,(r-c)!* sum(m=0,r-c+1,(-1)^(r-c+1-m)*m^r/m!/(r-c+1-m)!)))^-1)[n+1,k+1]
%Y Cf. A075263, A106340, A106339, A008278, A002206.
%K sign,tabl
%O 0,5
%A _Paul D. Hanna_, May 01 2005